hdu 2669 Romantic 扩展欧几里得

Now tell you two nonnegative integer a and b. Find the nonnegative integer X and integer Y to
satisfy X*a + Y*b = 1. If no such answer print "sorry" instead.

Input
The input contains multiple test cases.
Each case two nonnegative integer a,b (0<a, b<=2^31)

Output
output nonnegative integer X and integer Y, if there are more answers than the X smaller one
will be choosed. If no answer put "sorry" instead.

Sample Input
77 51
10 44
34 79

Sample Output
2 -3
sorry
7 -3

思路：exgcd 最后X为最小正解

代码：

 #include <bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 ll exgcd(ll a, ll b, ll &x, ll &y) {

     if(!b) {

         x=;y=;return a;

     }

     ll ans=exgcd(b,a%b,x,y);

     ll temp=x;

     x=y;

     y=temp-a/b*y;

     return ans;

 }

 int main() {

     ios::sync_with_stdio(false);

     ll a,b,x,y;

     while(cin>>a>>b) {

         ll g=exgcd(a,b,x,y);

         if(g!=) {

             cout<<"sorry"<<endl;

         } else {

             x=(x%b+b)%b;

             y=(-a*x)/b;

             cout<<x<<" "<<y<<endl;

         }

     }

     return ;

 }

巴特西

hdu 2669 Romantic 扩展欧几里得

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